Hyperpolarization

This week's Riddler Classic was too interesting to miss. Starting with vertically polarized light, how do you end up with horizontally polarized light? The answer of course is to use a polarizer oriented horizontally along with an intermediate polarizer at an eighth turn. The puzzle instead asks, assuming an arbitrary supply of polarizers with 1% reflectivity, how many should you use to get the optimal intensity out?

For the first polarizer offset at $\theta_1$, the resulting intensity would be $t\cos\theta_1$, where $t=0.99$ is the transmittance. So with $n$ polarizers we would have a final intensity of \begin{gather} I = \prod_i^n t \cos\theta_i \\ \text{with } \sum_i^n\theta_i = \frac\pi2 \end{gather}

Of course, if the polarizers had perfect transmittance for polarized light, i.e. $t=1$ then we would just want infinitely many lenses making small turns. Now, $\cos\theta$ is not a linear function, and even for small angles $$ \cos\theta \approx 1 - \frac{\theta^2}2 + \frac{\theta^4}{24} $$ but for $t=0.99$, there definitely will not be a win for varying $\theta_i$. Taking $$\theta = \theta_i = \theta_j \Longrightarrow \theta = \frac\pi{2n}~.$$ Then we have simply $$I = t^n \cos^n\theta = t^n\cos^n\left(\frac\pi{2n}\right) ~.$$

For $t=0.99$, this is maximized for $n=11$, outputting 80.042% of the original intensity.